In a deep drawing operation of rectangular sheet metal 30% stretching in length results in 15% reduction in thickness. Assuming volume constancy, the Normal anisotropy of sheet metal is _______ (Round off to two decimal places).

The correct answer is 0.625.

Here is the step-by-step explanation of the solution:

1. Define the parameters and relations:
In a sheet metal forming process, the normal anisotropy (Lankford coefficient, r) is defined as the ratio of the true strain in the width direction (${\epsilon }_w$) to the true strain in the thickness direction (${\epsilon }_t$):

$r=\frac{{\epsilon }_w}{{\epsilon }_t}$

2. Calculate the True Strains:
True strain ($\epsilon$) is related to engineering strain (e) by the formula:

$\epsilon =\ln (1+e)$

For the length direction, stretching is 30% ($e_l=0.30$):

${\epsilon }_l=\ln (1+0.30)=\ln \left(1.30\right)\approx 0.2624$

Rounding to two decimal places gives:

${\epsilon }_l\approx 0.26$

For the thickness direction, there is a 15% reduction ($e_t=-0.15$):

${\epsilon }_t=\ln (1-0.15)=\ln \left(0.85\right)\approx -0.1625$

Rounding to two decimal places gives:

${\epsilon }_t\approx -0.16$

3. Apply the Volume Constancy Condition:
Assuming plastic deformation is volume-constant, the sum of the true strains in all three principal directions must equal zero:

${\epsilon }_l+{\epsilon }_w+{\epsilon }_t=0$

Substituting the values of ${\epsilon }_l$ and ${\epsilon }_t$:

${\epsilon }_w=-({\epsilon }_l+{\epsilon }_t)$

${\epsilon }_w=-(0.26-0.16)=-0.10$

4. Calculate Normal Anisotropy (r):
Using the definition of normal anisotropy:

$r=\frac{{\epsilon }_w}{{\epsilon }_t}=\frac{-0.10}{-0.16}=0.625$